Đáp án:

$\begin{array}{l}
a)5\left( {x + 3} \right) - {x^2} - 3x = 0\\
 \Rightarrow 5\left( {x + 3} \right) - x\left( {x + 3} \right) = 0\\
 \Rightarrow \left( {x + 3} \right)\left( {5 - x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 5
\end{array} \right.\\
\text{Vậy}\,x =  - 3;x = 5\\
b)\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + x\left( {x + 1} \right)\left( {1 - x} \right) = 0\\
 \Rightarrow {x^3} - {3^3} + x\left( {1 - {x^2}} \right) = 0\\
 \Rightarrow {x^3} - 27 + x - {x^3} = 0\\
 \Rightarrow x = 27\\
\text{Vậy}\,x = 27\\
c)2{x^2} - \left( {x - 1} \right)\left( {x + 1} \right) = 2\\
 \Rightarrow 2{x^2} - \left( {{x^2} - 1} \right) = 2\\
 \Rightarrow 2{x^2} - {x^2} + 1 = 2\\
 \Rightarrow {x^2} = 1\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 1
\end{array} \right.\\
\text{Vậy}\,x = 1;x =  - 1\\
d){x^2} - 10 = 25\\
 \Rightarrow {x^2} = 35\\
 \Rightarrow \left[ \begin{array}{l}
x = \sqrt {35} \\
x =  - \sqrt {35} 
\end{array} \right.\\
\text{Vậy}\,x = \sqrt {35} ;x =  - \sqrt {35} \\
e)x\left( {2x - 1} \right)\left( {x + 5} \right) - \left( {2{x^2} + 1} \right)\left( {x + 4} \right) = 3,5\\
 \Rightarrow \left( {2{x^2} - x} \right)\left( {x + 5} \right) - \left( {2{x^3} + 8{x^2} + x + 4} \right) = 3,5\\
 \Rightarrow 2{x^3} + 10{x^2} - {x^2} - 5x - 2{x^3} - 8{x^2} - x - 4 = 3,5\\
 \Rightarrow {x^2} - 6x - 7,5 = 0\\
 \Rightarrow {x^2} - 6x + 9 - 16,5 = 0\\
 \Rightarrow {\left( {x - 3} \right)^2} = 16,5\\
 \Rightarrow \left[ \begin{array}{l}
x = \sqrt {16,5}  + 3\\
x =  - \sqrt {16,5}  + 3
\end{array} \right.\\
\text{Vậy}\,x =  \pm \sqrt {16,5}  + 3\\
f)\left( {3{x^2} - x + 1} \right)\left( {x - 1} \right) + {x^2}\left( {4 - 3x} \right) = \dfrac{5}{2}\\
 \Rightarrow 3{x^3} - 3{x^2} - {x^2} + x + x - 1 + 4{x^2} - 3{x^3} = \dfrac{5}{3}\\
 \Rightarrow 2x - 1 = \dfrac{5}{3}\\
 \Rightarrow 2x = \dfrac{8}{3}\\
 \Rightarrow x = \dfrac{4}{3}\\
\text{Vậy}\,x = \dfrac{4}{3}
\end{array}$